Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;
public class A
{
      public static void main(String ar[]) throws Exception
      {
            BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
            String s1[]=br.readLine().split(" ");
            int n=Integer.parseInt(s1[0]);
            int m=Integer.parseInt(s1[1]);
            
            int a[][]=new int[n][m];
            int b[][]=new int[n][m];
            StringBuffer sb=new StringBuffer();
            for(int i=0;i<n;i++)
            {
                  String s2[]=br.readLine().split(" ");
                  for(int j=0;j<m;j++)
                   a[i][j]=Integer.parseInt(s2[j]);
            }
            
            int r=0;
            if(n<=m)
            {
            for(int i=0;i<n;i++)
            {
                  int min=500;
                  for(int j=0;j<m;j++)
                   min=Math.min(min,a[i][j]-b[i][j]);
                  for(int j=0;j<m;j++)
                   b[i][j]+=min;
                  for(int j=0;j<min;j++)
                  { sb.append("row "+(i+1)).append("\n"); r++; }
                  //System.out.println(Arrays.toString(b[i]));
            }
            
            for(int j=0;j<m;j++)
            {
                  int min=500;
                  for(int i=0;i<n;i++)
                   min=Math.min(min,a[i][j]-b[i][j]);
                  for(int i=0;i<n;i++)
                   b[i][j]+=min;
                  for(int i=0;i<min;i++)
                  { sb.append("col "+(j+1)).append("\n");  r++; }
            }
            }
            else
            {
            
            for(int j=0;j<m;j++)
            {
                  int min=500;
                  for(int i=0;i<n;i++)
                   min=Math.min(min,a[i][j]-b[i][j]);
                  for(int i=0;i<n;i++)
                   b[i][j]+=min;
                  for(int i=0;i<min;i++)
                  { sb.append("col "+(j+1)).append("\n");  r++; }
            }  
            
                for(int i=0;i<n;i++)
            {
                  int min=500;
                  for(int j=0;j<m;j++)
                   min=Math.min(min,a[i][j]-b[i][j]);
                  for(int j=0;j<m;j++)
                   b[i][j]+=min;
                  for(int j=0;j<min;j++)
                  { sb.append("row "+(i+1)).append("\n"); r++; }
                  //System.out.println(Arrays.toString(b[i]));
            }
            }
            boolean bb=true;
            for(int i=0;i<n;i++)
             for(int j=0;j<m;j++)
              if(a[i][j]!=b[i][j])
               bb=false;
            
            if(!bb)
             System.out.println("-1");
            else
            {
              System.out.println(r);
              System.out.println(sb);
            }
      }
}